Chapter 4 - Types of Chemical Reactions and Solution Stoichiometry - Exercises - Page 181: 39

4.5 M

Sodium carbonate: $Na_2CO_3\Rightarrow 2Na^++CO_3^{2-}$ Sodium bicarbonate: $NaHCO_3\Rightarrow Na^{+}+HCO_3^{-}$ Moles of sodium ion = $0.070 \text{ L}*3.0\text{ M}*2+0.030 \text{ L}*1.0\text{ M}*1=0.45$ moles Total liters of solution = 0.0700 L + 0.0300 L = 0.100 L Sodium ion concentration = $\frac{0.45 \text{ moles}}{0.100 \text{ L}}=4.5\text{ M}$